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The perimeter of a rectangular playing field is 244 feet. If its length is 2 feet longer than twice its width, what are the dimensions of the field?

A. 20 ft./41 ft.
B. 21 ft./40 ft.
C. 40 ft./82 ft.
D. 42 ft./80 ft.

User Ooak
by
6.6k points

2 Answers

3 votes

Final answer:

To find the dimensions of the playing field, we used the given perimeter and the relationship between the length and width. Solving the resulting equation shows that the field's dimensions are 40 feet in width and 82 feet in length.

Step-by-step explanation:

To determine the dimensions of a rectangular playing field with a perimeter of 244 feet and a length that is 2 feet longer than twice its width, we need to set up equations based on the given information. Let's denote the width of the field as w and the length as l.

According to the problem, l = 2w + 2. The formula for the perimeter P of a rectangle is P = 2l + 2w. Substituting the expression for l into the perimeter formula we get P = 2(2w + 2) + 2w. Since the perimeter is 244 feet, we can write this as 244 = 4w + 4 + 2w, which simplifies to 244 = 6w + 4.

Moving the constant to the left side gives us 240 = 6w, and dividing both sides by 6 yields w = 40. Given that l = 2w + 2, we can find the length by substituting w into this equation, giving us l = 2(40) + 2 = 80 + 2 = 82. Therefore, the dimensions of the field are width = 40 feet and length = 82 feet, which makes the answer (C) 40 ft./82 ft.

User Muhammet Ali Asan
by
6.2k points
3 votes
advice : use cy math.com on your computer
or photo math on your phone
User Jesse Cohen
by
6.9k points
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